For the following reaction mechanism: 1) A + B [k_-1]k_1 I (fast equilibrium, K_1=k_1/(k_-1)) 2) I + C [k_-2]k_2 J (fast equilibrium, K_2=k_2/(k_-2)) 3) J k_3 P (slow)

For the following reaction mechanism: 1) $A + B k_{1} k_{- 1} I$ (fast equilibrium, $K_{1} = \frac{k _{1}}{k _{- 1}}$ ) 2) $I + C k_{2} k_{- 2} J$ (fast equilibrium, $K_{2} = \frac{k _{2}}{k _{- 2}}$ ) 3) $J k_{3} P$ (slow)

hard 5 marks Class 12 · Chemistry · Chemical Kinetics

Answer

(a)

r = k_{3} K_{1} K_{2} [A] [B] [C]

; (b) first order in

A

B

and

C

(overall third order); (c) Step 3 is the rate‑determining step because it is the only slow step, and the fast pre‑equilibria adjust instantly to maintain equilibrium concentrations.

Step-by-step solution

The overall rate equals the rate of the slow step: $r = k_{3} [J]$ .
From the second fast equilibrium: $K_{2} = \frac{[ J ]}{[ I ] [ C ]} \Rightarrow [J] = K_{2} [I] [C]$ .
From the first fast equilibrium: $K_{1} = \frac{[ I ]}{[ A ] [ B ]} \Rightarrow [I] = K_{1} [A] [B]$ .
Substitute $[I]$ into the expression for $[J]$ : $[J] = K_{2} (K_{1} [A] [B]) [C] = K_{1} K_{2} [A] [B] [C]$ .
Insert $[J]$ into the rate expression: $r = k_{3} K_{1} K_{2} [A] [B] [C]$ .
Each reactant appears to the first power, so the reaction is first order in $A$ , first order in $B$ , first order in $C$ , giving an overall order of three.
Step 3 is marked as slow; the fast pre‑equilibria (steps 1 and 2) rapidly establish concentrations of $I$ and $J$ , therefore the overall reaction rate is controlled by the slow conversion of $J$ to $P$ .

Common mistake: Treating the second step as the rate‑determining step and forgetting to express

[J]

through both equilibrium constants.

Concept tested: Rate law from a mechanism with pre‑equilibrium

Published 14 September 2026 · How this solution was prepared and checked

← All Chemical Kinetics questions

For the following reaction mechanism: 1) A+Bk1​k−1​​I (fast equilibrium, K1​=k−1​k1​​) 2) I+Ck2​k−2​​J (fast equilibrium, K2​=k−2​k2​​) 3) Jk3​​P (slow)

Step-by-step solution

For the following reaction mechanism: 1) $A + B k_{1} k_{- 1} I$ (fast equilibrium, $K_{1} = \frac{k _{1}}{k _{- 1}}$ ) 2) $I + C k_{2} k_{- 2} J$ (fast equilibrium, $K_{2} = \frac{k _{2}}{k _{- 2}}$ ) 3) $J k_{3} P$ (slow)